Is the notion of irreducubility of polynomials defined only over fields, or can it be any set?
On Wikipedia, it only talks about a polynomial being irreducible over a field. But what if I want to talk about, say, a quadratic with integer coefficients whose factors consist of non-integers. Can I still say that the polynomial is irreducible over $ mathbb{Z} $, even though $ mathbb{Z} $ does not constitute a field?
Or how about if I want to talk about a polynomial whose coefficients belong to some set $ S $, where $ S $ is not a field, having factors with coefficients that do not belong to $ S $. Can I still use the term 'irreducible'?
irreducible-polynomials
asked Dec 12 '18 at 23:22
A.AbbasA.Abbas
296
add a comment |
On Wikipedia, it only talks about a polynomial being irreducible over a field. But what if I want to talk about, say, a quadratic with integer coefficients whose factors consist of non-integers. Can I still say that the polynomial is irreducible over $ mathbb{Z} $, even though $ mathbb{Z} $ does not constitute a field?
Or how about if I want to talk about a polynomial whose coefficients belong to some set $ S $, where $ S $ is not a field, having factors with coefficients that do not belong to $ S $. Can I still use the term 'irreducible'?
irreducible-polynomials
asked Dec 12 '18 at 23:22
A.AbbasA.Abbas
296
1
Where are you seeing this? The first paragraph of the article on irreducible polynomials talks about irreducibility over $mathbb{Z}$. Additionally, the second paragraph in the "definition" section talks about extending the definition to unique factorization domains.
– platty
Dec 12 '18 at 23:23
add a comment |
On Wikipedia, it only talks about a polynomial being irreducible over a field. But what if I want to talk about, say, a quadratic with integer coefficients whose factors consist of non-integers. Can I still say that the polynomial is irreducible over $ mathbb{Z} $, even though $ mathbb{Z} $ does not constitute a field?
Or how about if I want to talk about a polynomial whose coefficients belong to some set $ S $, where $ S $ is not a field, having factors with coefficients that do not belong to $ S $. Can I still use the term 'irreducible'?
irreducible-polynomials
asked Dec 12 '18 at 23:22
A.AbbasA.Abbas
296
On Wikipedia, it only talks about a polynomial being irreducible over a field. But what if I want to talk about, say, a quadratic with integer coefficients whose factors consist of non-integers. Can I still say that the polynomial is irreducible over $ mathbb{Z} $, even though $ mathbb{Z} $ does not constitute a field?
Or how about if I want to talk about a polynomial whose coefficients belong to some set $ S $, where $ S $ is not a field, having factors with coefficients that do not belong to $ S $. Can I still use the term 'irreducible'?
irreducible-polynomials
irreducible-polynomials
asked Dec 12 '18 at 23:22
A.AbbasA.Abbas
296
asked Dec 12 '18 at 23:22
A.AbbasA.Abbas
296
asked Dec 12 '18 at 23:22
A.AbbasA.Abbas
296
asked Dec 12 '18 at 23:22
A.AbbasA.Abbas
296
asked Dec 12 '18 at 23:22
A.AbbasA.Abbas
296
296
1
Where are you seeing this? The first paragraph of the article on irreducible polynomials talks about irreducibility over $mathbb{Z}$. Additionally, the second paragraph in the "definition" section talks about extending the definition to unique factorization domains.
– platty
Dec 12 '18 at 23:23
add a comment |
1
Where are you seeing this? The first paragraph of the article on irreducible polynomials talks about irreducibility over $mathbb{Z}$. Additionally, the second paragraph in the "definition" section talks about extending the definition to unique factorization domains.
– platty
Dec 12 '18 at 23:23
1
1
Where are you seeing this? The first paragraph of the article on irreducible polynomials talks about irreducibility over $mathbb{Z}$. Additionally, the second paragraph in the "definition" section talks about extending the definition to unique factorization domains.
– platty
Dec 12 '18 at 23:23
Where are you seeing this? The first paragraph of the article on irreducible polynomials talks about irreducibility over $mathbb{Z}$. Additionally, the second paragraph in the "definition" section talks about extending the definition to unique factorization domains.
– platty
Dec 12 '18 at 23:23
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037369%2fis-the-notion-of-irreducubility-of-polynomials-defined-only-over-fields-or-can%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037369%2fis-the-notion-of-irreducubility-of-polynomials-defined-only-over-fields-or-can%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Where are you seeing this? The first paragraph of the article on irreducible polynomials talks about irreducibility over $mathbb{Z}$. Additionally, the second paragraph in the "definition" section talks about extending the definition to unique factorization domains.
– platty
Dec 12 '18 at 23:23